Finding Exact Values of Trig Expressions w/o Calculator

[GreenPrint]

Homework Statement

Use properties of the trigonometric functions to find the exact value of each expression. Do not use a calculator

sin (-pi/12) csc (25 pi)/12

Homework Equations

sin (negative angle) = - sin (angle)
csc theta = (sin theta)^-1 = 1/(sin theta)

The Attempt at a Solution

Ok there is obviously some sort of property of trig identies I do not know so I'm struggling here...

sin(-pi/12) csc( (25 pi)/12 ) = (- sin(pi/12) )/(sin( (25 pi)/12))
know if I'm correct I just treat them as exponents correct? so I subtract them

- sin( (pi/12) - ( (25 pi)/12 )
- sin( (-24 pi)/12 )
+ sin( (24 pi)/12
sin 2 pi = 0

OK THIS IS WRONG I put this into my calculator and i get negative one. I postulated for like two hours on how to do this. I think I figuered it out but am not sure why it works can someone please explain it to me...

ok I start out here
sin(-pi/12) csc( (25 pi)/12 ) = (- sin(pi/12) )/(sin( (25 pi)/12))


then
- sin( (pi/12) - ( (25 pi)/12 )

- sin( (24 pi)/12 )

if i just ignore the negative sign next to the 24... something here is wrong with what I'm doing HELP

- sin( (24 pi)/12 )

now I take the recipical but leave the pi on top of the angle not sure why or why i leave pi on top

- sin( (12 pi)/24

simplify

- sin( pi/2

pi/2 is ninety degrees which has the coordinates (0,1) and sense sine is equal to the y cordinate I get one but sense it's the opposite i get negative 1 as my answer which is correct

So I obviously have no idea how to really do this problem if someone could tell me how to do this that would be great. There isn't really much I can do because I'm obviously don't know some property or something here.

Thanks

 

[LCKurtz]

None of your algebraic manipulations make any sense. You are just making up operations that don't exist.

Notice that Pi/12 is half of pi/6 and you know the functions for multiples of pi/6. Look at the half angle formulas instead of trying to make up your own.

 

[GreenPrint]

I have yet to have been taught these formulas yet so there must be some other way to do it

 

[Anakin_k]

Here are the two special triangles I've used in the past: http://fouss.pbworks.com/f/special triangle 3.JPG and http://fouss.pbworks.com/f/special triangle 2.JPG and recall that sin (a-b)=(sin a)(cos b)-(sin b)(cos a)

Now sin $$\frac{-pi}{12}$$ = sin ($$\frac{pi}{6}$$ - $$\frac{pi}{4}$$) = sin $$\frac{pi}{6}$$ * cos $$\frac{pi}{4}$$ - sin $$\frac{pi}{4}$$ * cos $$\frac{pi}{6}$$

Use the special triangles (unless you have them memorized, which you should have) and solve.

Edit: Sorry I thought you were doing two different equations. I now see that sin (-pi/12) * csc (25 pi)/12 is what you want. I guess you can do the second one and after solving it, multiple both answers.